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#10510: Fix bug in forward port of 2.7 distutils patch.
Pointed out by Arfrever.
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3 changed files with 77 additions and 3 deletions
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@ -401,7 +401,7 @@ requested again. This is called "memoizing", and can be implemented like this::
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# Calculate the value
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# Calculate the value
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result = ... expensive computation ...
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result = ... expensive computation ...
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_cache[(arg1, arg2)] = result # Store result in the cache
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_cache[(arg1, arg2)] = result # Store result in the cache
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return result
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return result
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You could use a global variable containing a dictionary instead of the default
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You could use a global variable containing a dictionary instead of the default
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@ -448,6 +448,81 @@ arguments a function can accept. For example, given the function definition::
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the values ``42``, ``314``, and ``somevar`` are arguments.
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the values ``42``, ``314``, and ``somevar`` are arguments.
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Why did changing list 'y' also change list 'x'?
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------------------------------------------------
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If you wrote code like::
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>>> x = []
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>>> y = x
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>>> y.append(10)
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>>> y
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[10]
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>>> x
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[10]
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you might be wondering why appending an element to ``y`` changed ``x`` too.
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There are two factors that produce this result:
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1) Variables are simply names that refer to objects. Doing ``y = x`` doesn't
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create a copy of the list -- it creates a new variable ``y`` that refers to
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the same object ``x`` refers to. This means that there is only one object
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(the list), and both ``x`` and ``y`` refer to it.
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2) Lists are :term:`mutable`, which means that you can change their content.
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After the call to :meth:`~list.append`, the content of the mutable object has
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changed from ``[]`` to ``[10]``. Since both the variables refer to the same
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object, accessing either one of them accesses the modified value ``[10]``.
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If we instead assign an immutable object to ``x``::
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>>> x = 5 # ints are immutable
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>>> y = x
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>>> x = x + 1 # 5 can't be mutated, we are creating a new object here
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>>> x
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6
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>>> y
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5
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we can see that in this case ``x`` and ``y`` are not equal anymore. This is
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because integers are :term:`immutable`, and when we do ``x = x + 1`` we are not
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mutating the int ``5`` by incrementing its value; instead, we are creating a
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new object (the int ``6``) and assigning it to ``x`` (that is, changing which
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object ``x`` refers to). After this assignment we have two objects (the ints
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``6`` and ``5``) and two variables that refer to them (``x`` now refers to
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``6`` but ``y`` still refers to ``5``).
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Some operations (for example ``y.append(10)`` and ``y.sort()``) mutate the
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object, whereas superficially similar operations (for example ``y = y + [10]``
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and ``sorted(y)``) create a new object. In general in Python (and in all cases
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in the standard library) a method that mutates an object will return ``None``
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to help avoid getting the two types of operations confused. So if you
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mistakenly write ``y.sort()`` thinking it will give you a sorted copy of ``y``,
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you'll instead end up with ``None``, which will likely cause your program to
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generate an easily diagnosed error.
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However, there is one class of operations where the same operation sometimes
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has different behaviors with different types: the augmented assignment
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operators. For example, ``+=`` mutates lists but not tuples or ints (``a_list
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+= [1, 2, 3]`` is equivalent to ``a_list.extend([1, 2, 3])`` and mutates
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``a_list``, whereas ``some_tuple += (1, 2, 3)`` and ``some_int += 1`` create
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new objects).
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In other words:
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* If we have a mutable object (:class:`list`, :class:`dict`, :class:`set`,
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etc.), we can use some specific operations to mutate it and all the variables
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that refer to it will see the change.
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* If we have an immutable object (:class:`str`, :class:`int`, :class:`tuple`,
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etc.), all the variables that refer to it will always see the same value,
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but operations that transform that value into a new value always return a new
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object.
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If you want to know if two variables refer to the same object or not, you can
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use the :keyword:`is` operator, or the built-in function :func:`id`.
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How do I write a function with output parameters (call by reference)?
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How do I write a function with output parameters (call by reference)?
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---------------------------------------------------------------------
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---------------------------------------------------------------------
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@ -164,7 +164,6 @@ class upload(PyPIRCCommand):
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if value and value[-1:] == b'\r':
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if value and value[-1:] == b'\r':
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body.write(b'\n') # write an extra newline (lurve Macs)
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body.write(b'\n') # write an extra newline (lurve Macs)
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body.write(end_boundary)
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body.write(end_boundary)
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body.write(b"\r\n")
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body = body.getvalue()
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body = body.getvalue()
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self.announce("Submitting %s to %s" % (filename, self.repository), log.INFO)
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self.announce("Submitting %s to %s" % (filename, self.repository), log.INFO)
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@ -127,7 +127,7 @@ class uploadTestCase(PyPIRCCommandTestCase):
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# what did we send ?
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# what did we send ?
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headers = dict(self.last_open.req.headers)
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headers = dict(self.last_open.req.headers)
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self.assertEqual(headers['Content-length'], '2163')
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self.assertEqual(headers['Content-length'], '2161')
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content_type = headers['Content-type']
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content_type = headers['Content-type']
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self.assertTrue(content_type.startswith('multipart/form-data'))
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self.assertTrue(content_type.startswith('multipart/form-data'))
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self.assertEqual(self.last_open.req.get_method(), 'POST')
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self.assertEqual(self.last_open.req.get_method(), 'POST')
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